Question: Let $h(x)=\sqrt{x^2+4}$ Where does $h$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-2$ (Choice B) B $x=0$ (Choice C) C $x=2$ (Choice D) D $h$ has no critical points.
Explanation: A critical point of $h$ is a point in the domain of $h$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $h$, let's find its derivative. $\begin{aligned} h'(x)&=\dfrac{d}{dx}\left[ \sqrt{x^2+4} \right] \\\\ &=\dfrac{1}{2\sqrt{x^2+4}}\cdot \dfrac{d}{dx}[x^2+4] \\\\ &=\dfrac{2x}{2\sqrt{x^2+4}} \\\\ &=\dfrac{x}{\sqrt{x^2+4}} \end{aligned}$ Now let's look for $x$ -values where $h'$ is zero or undefined. $\dfrac{x}{\sqrt{x^2+4}}=0$ at $x=0$. $\dfrac{x}{\sqrt{x^2+4}}$ is never undefined, so $h'$ is never undefined. In conclusion, this is the only $x$ -value where $h$ has a critical point: $x=0$